For each positive real number $x$, let $A_x$ be the set of positive real numbers $y$ satisfying $(x + f(y))(y + f(x)) \le 4$. The following are easy consequences by the definition.
(1) If $x \in A_y$, then $y \in A_x$.
(2) For $x < y$, if $f(x) \le f(y)$ then $A_y \subseteq A_x$.
We prove the following lemma.
Lemma. For $x \in \mathbb{R}^+$, there are only finitely many positive real numbers $y (< x)$ such that $f(y) < f(x)$.
Proof. Since $A_x$ is not empty, there exists $z \in A_x$. For $y (< x)$ with $f(y) < f(x)$, we have $A_x \subseteq A_y$ by (2). So $z$ also belongs to $A_y$, and $y \in A_z$ by (1). Therefore, if there are infinitely many $y (< x)$ such that $f(y) < f(x)$, $A_z$ becomes an infinite set, which is a contradiction. $\square$
Assume that there exist $a, b \in \mathbb{R}^+$ such that $a < b$ and $f(a) \le f(b)$. Let $S = \{x \mid a < x < b, f(x) > f(a)\}$. By the lemma, there are only finitely many $x$ with $a < x < b$ such that $f(x) < f(b)$. So $S$ is an infinite set. For any $s \in S$, $A_s \subseteq A_a$ by (2). Since $A_s$ is not empty, there exists $t \in A_s \subseteq A_a$. So, $s \in A_t \subseteq \bigcup_{z \in A_a} A_z$ by (1). That is, $S \subseteq \bigcup_{z \in A_a} A_z$ which yields a contradiction because $S$ is an infinite set but
$$| \bigcup_{z \in A_a} A_z | \le \sum_{z \in A_a} |A_z|$$
is finite. $\square$